1895.] The Latent Heat of Evaporation of Water, 219 



This must have been (as Regnault himself points out) con- 

 siderably below the pressure of the vapour in the flask. 



(2.) At low temperatures a small error in the pressure would 

 cause a considerable error in temperature. For example, 

 an error of 0*4 mm. in pressure would correspond (at 4 C.) 

 to an error of 1 in 0. 



(3.) All his observations of the change in temperature of the 

 calorimeter at low temperatures were taken on a falling 

 mercury thermometer, and an error of 0'01 (the limit to 

 which he observed his thermometer) would change the 

 resulting value of L by 1 in 500, Now, no observations on 

 falling mercury thermometers can be relied 011 to give an 

 accuracy of anything like the above order. 



(4.) The extreme divergence between the results of these experi- 

 ments would alone render them of small value. 



(5.) Regnault's formula for the "total heat" (606'5 + 0'305 0) is 

 not in agreement with his experimental results. 



I have in my paper given further reasons for their rejection, but 

 T think the above are alone sufficient to prove that at any rate the 

 results of these experiments must be regarded with suspicion. I have 

 ulso shown that the tendency of all the above sources of error is to 

 make Regnault's values of L too high at these temperatures. 



These above objections, however, lose all their force when applied 

 to Regnault's determinations from 63 to 100. His methods of ex- 

 periment were entirely altered ; for example, p was observed where 

 x .he vapour was formed, and a large change in p would produce but 

 =a small change in 0. Also his thermometers were rising, the range of 

 temperature was more than twice as great, and his results were far 

 more uniform. &c. 



At about 100 Regnault performed forty-four experiments, of 

 which he rejects the first six as "preliminary "; the remaining thirty- 

 ight give the following mean results : 



Temp. " Total heat." 



99-88 636-67 



This would become 636'60 at 100. Hence, if we assume that 

 1 gram of water in cooling from 100 to gives out 100 thermal 

 units, we get L = 536 '60 at 100. 



Now, taking my values, viz., 



Temp. L. 



40-15 572-60 



30-00 578-70 



We get dl^jde = 0'6010. 



R 2 



